Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $q = \dfrac{a + 5}{10a - 10} \div \dfrac{a^2 + 5a}{a^2 - 3a + 2} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{a + 5}{10a - 10} \times \dfrac{a^2 - 3a + 2}{a^2 + 5a} $ First factor the quadratic. $q = \dfrac{a + 5}{10a - 10} \times \dfrac{(a - 1)(a - 2)}{a^2 + 5a} $ Then factor out any other terms. $q = \dfrac{a + 5}{10(a - 1)} \times \dfrac{(a - 1)(a - 2)}{a(a + 5)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (a + 5) \times (a - 1)(a - 2) } { 10(a - 1) \times a(a + 5) } $ $q = \dfrac{ (a + 5)(a - 1)(a - 2)}{ 10a(a - 1)(a + 5)} $ Notice that $(a + 5)$ and $(a - 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ \cancel{(a + 5)}(a - 1)(a - 2)}{ 10a\cancel{(a - 1)}(a + 5)} $ We are dividing by $a - 1$ , so $a - 1 \neq 0$ Therefore, $a \neq 1$ $q = \dfrac{ \cancel{(a + 5)}\cancel{(a - 1)}(a - 2)}{ 10a\cancel{(a - 1)}\cancel{(a + 5)}} $ We are dividing by $a + 5$ , so $a + 5 \neq 0$ Therefore, $a \neq -5$ $q = \dfrac{a - 2}{10a} ; \space a \neq 1 ; \space a \neq -5 $